A 290-m-wide river flows due east at a uniform speed of 2.9 m/s. A boat with a s

Question

A 290-m-wide river flows due east at a uniform speed of 2.9 m/s. A boat with a speed of 7.3 m/s relative to the water leaves the south bank pointed in a direction 32o west of north. What is the (a) magnitude and (b) direction of the boat's velocity relative to the ground? Give the direction as the angle of the velocity from due north, positive if to the east and negative if to the west. (c) How long does it take for the boat to cross the river?

I'm actually just stumped on this, and don't know how really to approach this.
any help will be much appreciated

Explanation / Answer

it is all in defining the subscripts

let r= river, g=ground, b=boat
then
vrg = 2.9 m/s i

vbr = -(7.3 m/s)cos 32 i + (7.3 m/s) sin 32 j

now

vbg = vbr + vrg

vbg = -(7.3 m/s)cos 32 i + (7.3 m/s) sin 32 j + 2.9 m/s i

vbg = (-7.3 m/s)cos 32+ 2.9 m/s) i + (7.3 m/s) sin 32 j

vbg = -3.29 m/s i + 3.87 m/s j

a) the magnitude is

vbg = sqrt(3.292 + 3.872)

vbg = 5.08 m/s

b) the direction

= tan-1( 3.87 / 3.29)

= -49.6 degrees (note it is negative because it is due west)

c) how long does it take to go 290 m

v=d/t so t = d/v

t = (290 m) / (5.08 m/s)

t = 57.1 s

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